Question

∆H_o = −3310 kJ/mol for: 4 FeS₂(s) + 11 O₂(g) → 2 Fe₂O₃(s) + 8 SO₂(g)

If ∆H^o_f [Fe₂O₃(s)] = -824.2 and ∆H^o_f [SO₂(g)] = −296.83 kJ/mol determine ∆H_o for Fe(s) + 2S(s) → FeS₂(s).

Answer 1

If 4 FeS2(s) + 11 O2(g) → 2 Fe2O3(s) + 8 SO2(g) and ∆H^orxn = -3310 kJ/mol

We know,

∆H^orxn = (sum of enthalpy of formation of product) - (sum of enthalpy of formation of reactant)

-3310 = ((-2×824.2)+(-8×296.83)) - (4×enthalpy of formation of FeS2 + 0)

Enthalpy 0f formation of FeS2 = (3310+(-4023.04))/4 = -178.26 kJ/mol

The second reaction is

Fe(s) + 2S(s) → FeS2(s)

∆H^orxn = (sum of enthalpy of formation of product) - (sum of enthalpy of formation of reactant)

= -178.26 - 0 = -178.26 kJ/mol

The enthalpy of formation of elemental molecules is always considered to be zero.

So the answer is -178.26 kJ/mol

While calculating I have not shown unit as all the energies are in kJ/mol.